Question

A particle of mass

m

and charge

q

is moving in a region where uniform, constant electric and magnetic

\vec{E}

&

\vec{B}

fields are present,

\vec{E}

&

\vec{B}

are parallel to each other. At time

t = 0

the velocity of the particle is perpendicular to

\vec{E}

. (Assume that its speed is always

< < c

, the speed of light in vacuum). Find the velocity

\vec{v}

of the particle at time

t

. You must express your answer in terms of

t, q, m,

the vectors

\vec{v_{0}}

,

\vec{E}

&

\vec{B}

and their magnitudes

v_{0}, E

and

B

.

Accepted Answer

Let us define the unit vectors: $\hat{j} = \frac{\vec{E}}{E}$ or $\frac{\vec{B}}{B}$ , $\hat{i} = \frac{\vec{v_{o}}}{v_{o}}$ and $\hat{k} = \frac{\vec{v_{o}} \times \vec{B}}{v_{o} B}$

Let as suppose initially particle is moving in x- direction as shown in the figure.

Force due to electric field will be along y-axis. Also, magnetic force will not affect the motion of charged particle in the direction of electric field (or y-axis here). Hence, acceleration of charge particle along y-direction,

$a_{y} = \frac{F_{E}}{m} = \frac{q E}{m}$ (Constant)

By using kinematic equation of motion along y-direction, y-component of velocity, $v_{y}$ can be given as,

$v_{y} = 0 + a_{y} t \Rightarrow v_{y} = \frac{q E}{m} t . . . (1)$

The charge particle under the action of magnetic field describe a circle in $X - Z$ plane at an angular speed $ω = \frac{q B}{m}$ in the plane perpendicular to $\vec{B}$ .

Let the position vector of charged particle $\vec{r}$ makes an angle $θ$ with z-axis then on resolving velocity $\vec{v_{o}}$ along x-axis and z-axis we get $v_{x} = v_{o} \cos (θ)$ and $v_{z} = v_{o} \sin (θ)$ as shown in figure below,

here, $θ = ω t \Rightarrow θ = \frac{q B}{m} t$ .

On substituting $θ$ in expression of $v_{x}$ and $v_{z}$ we get, $v_{x} = v_{o} \cos (\frac{q B}{m} t) . . . (2)$ and

$v_{z} = v_{o} \sin (\frac{q B}{m} t) . . . (3)$

Therefore instantaneous velocity $\vec{v}$ at time $t$ is given by $\vec{v} = v_{x} \hat{i} + v_{y} \hat{j} + v_{z} \hat{k}$ or

$\vec{v} =$ $v_{o} \cos (\frac{q B}{m} t)$ $(\frac{\vec{v_{o}}}{v_{o}})$ $+$ $\frac{q E}{m} t (\frac{\vec{E}}{E})$ $+$ $v_{o} \sin (\frac{q B}{m} t) (\frac{\vec{v_{o}} \times \vec{B}}{v_{o} B})$

Embibe Experts Solutions for Chapter: Magnetic Effect of Current, Exercise 4: Exercise - 4

Embibe Experts Physics Solutions for Exercise - Embibe Experts Solutions for Chapter: Magnetic Effect of Current, Exercise 4: Exercise - 4

Questions from Embibe Experts Solutions for Chapter: Magnetic Effect of Current, Exercise 4: Exercise - 4 with Hints & Solutions

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